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An exercise in my class is asking me to prove the following

Let P ∈ Lin(H). The following are equivalent :

(a) P is positive, i.e. $⟨ψ|P|ψ⟩ ≥ 0$ f or all $|ψ⟩ ∈ H$

(b) P is Hermitian all eigenvalues are non − negative.

(c) There exists $B ∈ Lin(\mathcal{H, K})$ such that $P = B^† B$ for some Hilbert space $\mathcal{K}$

(d) For every $Q ∈ PSD(H)$ we have $tr[PQ] ≥ 0$

What i have so far is:

Assuming (a) to be true we know that operator P is positive $⟨ψ|P|ψ⟩ ≥ 0$. This tells us that the result of the expression $⟨ψ|P|ψ⟩$ must be some real number bigger than or equal to $0$, e.g. $5$. From this we know that the following is true since one might replace the expression with any real number bigger than or equal to $0$, leaving the conjugate redundant as it only affects complex numbers

$$⟨ψ|P|ψ⟩^∗ = ⟨ψ|P|ψ⟩$$

which tells us that

$$⟨ψ|P|ψ⟩^∗ ≥ 0$$

With $⟨ψ|P|ψ⟩^∗$ being the conjugate and not the adjoint. From here off i am currently stuck and don't really know where to go in order to continue the proof. My current goal is to prove (b) by assuming (a) is true. I have read the post Show that a positive operator is also hermitian, but unfortunately this didn't help me. Any tips or help would be wonderful on how to prove (b), but also on how to go forward with proving (c), (b) and (a) when i get to them.

The task gives the tip "consider the matrix $M = i(P − P^†)$. What special property does $M$ have?", where $P^†$ is the adjoint. I do not understand the purpose of the hint though, most likely because i lack the intuition as this is my first quantum information course.

This is my first post so i do hope i upheld the guidelines of how to write questions here, but if i didn't i apoligise up ahead.

Mikkel
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  • For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Another User Feb 09 '23 at 14:09
  • Hint: You already found out that $\langle \psi|P|\psi\rangle^* = \langle \psi|P|\psi\rangle$, now use the general identity $\langle \psi|P|\psi\rangle^* = \langle \psi|P^\dagger|\psi\rangle$ – StiftungWarentest Feb 09 '23 at 14:49
  • @StiftungWarentest I am not familiar with this identity and unfortnately fails to see why the conjugate expression is equal to original expression but with the adjoint $P^†$. Can you elaborate on this? Thanks a lot for the help! – Mikkel Feb 09 '23 at 15:02
  • $\langle \psi|P^\dagger|\psi\rangle = \langle P\psi|\psi\rangle = \langle\psi|P\psi\rangle^* = \langle\psi|P|\psi\rangle^*$ – StiftungWarentest Feb 09 '23 at 15:24

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